Engineering Finding current/impedance in circuit

[johndough999]

Homework Statement

Homework Equations

Law of Superposition
I=V/R

The Attempt at a Solution

I am reducing the circuit.
i) By law of Superposition, short circuit V2
100 ohms on the leftmost branch and 50 ohms on the rightmost branch, both are in parallel so
100*80/(100+80)=44.4 ohms seen by V1(t)

ii)
I=V/R
I=2/44.4
I=0.045 Aiii) short circuit V1
reduce the circuit,
60+40+80=180 ohms as all the resistors are in series.

iv)
I=V/R
I=10/180
I=1/18 A

v)Using law of super position
By shorting V2 I obtain I=0.02A
By shorting V1 I obtain I= 1/18A
I = 0.02+1/18 = 0.076A

Please correct me where I am going wrong. Your help is appreciated. Thanks!

 

[Hesch]

Say there is 0V at the bottom. Then there must be V1 at the top: I_60+40Ω = V1 / R_60+40Ω.

The voltage over 80Ω = V1-V2. Calculate I_80Ω.

Now you can calculate the current in the leftmost branch by KCL.

 

[Hesch]

(iii) is wrong: Z_seen = 80Ω + ((60Ω+40Ω) || Z_V1). Thus (iv) is also wrong.

 

[johndough999]

For part iii) do I not have to short circuit V1? hence would i not be left with a series circuit?

 

[Hesch]

For part iii) do I not have to short circuit V1?

Yes, you may short circuit V1, but this is not necessary because it is already "short circuited" as to impedance. It is 0Ω as well as for V2.

Contrary the impedance in a current source is infinite.

 

[johndough999]

Zseen = 80Ω + ((60Ω+40Ω) || ZV1)

From the formula you mentioned does that mean 80Ω is in parallel with 60 and 40 Ω?

 

[Hesch]

Zseen = 80Ω + ((60Ω+40Ω) || ZV1)

By " || " , parallel is meant.

ZV1 = 0, so Zseen = 80Ω + (100Ω || 0Ω) = 80Ω

 

[donpacino]

Homework Statement

View attachment 83092

Homework Equations

Law of Superposition
I=V/R

The Attempt at a Solution

I am reducing the circuit.
i) By law of Superposition, short circuit V2
100 ohms on the leftmost branch and 50 ohms on the rightmost branch, both are in parallel so
100*80/(100+80)=44.4 ohms seen by V1(t)

ii)
I=V/R
I=2/44.4
I=0.045 Aiii) short circuit V1
reduce the circuit,
60+40+80=180 ohms as all the resistors are in series.

iv)
I=V/R
I=10/180
I=1/18 A

v)Using law of super position
By shorting V2 I obtain I=0.02A
By shorting V1 I obtain I= 1/18A
I = 0.02+1/18 = 0.076A

Please correct me where I am going wrong. Your help is appreciated. Thanks!

in parts 3 and 4, what is the voltage across the 60 and 40 ohm resistors?
recall ohms law, what does that tell you about the current through the resistors.

 

[johndough999]

in parts 3 and 4, what is the voltage across the 60 and 40 ohm resistors?
recall ohms law, what does that tell you about the current through the resistors.

donpacino could you elaborate a little more. Would the voltage across the 60 and 40 ohm be the total voltage - voltage across 80 ohm?.

 

[donpacino]

donpacino could you elaborate a little more. Would the voltage across the 60 and 40 ohm be the total voltage - voltage across 80 ohm?.

in parts 3 and 4 you said you were setting V1 to zero... If you set V1 to zero, what is the voltage across the 60 and 40 ohm resistors

 

[johndough999]

in parts 3 and 4 you said you were setting V1 to zero... If you set V1 to zero, what is the voltage across the 60 and 40 ohm resistors

the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor

 

[donpacino]

the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor

nope. V1 is short circuited, which means the voltage across it is zero

 

[donpacino]

the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor

well, technically that is correct, what is the voltage across the 80 ohm resistor?

 

[johndough999]

So if V1 is short circuited and the voltage across it zero, voltage across 60 and 40 ohms is zero therefore voltage across 80 ohm resistor is equal to V2?

 

[donpacino]

So if V1 is short circuited and the voltage across it zero, voltage across 60 and 40 ohms is zero therefore voltage across 80 ohm resistor is equal to V2?

yes. so knowing that, look at your answer to parts 3 and 4 again

 

[johndough999]

iii) 80 0hms

iv) I=V/R
I=10/80
I=1/8 A

v) 0.02A? since current in the centre branch when V1 is short circuited is 0A.

 

[donpacino]

iii) 80 0hms

iv) I=V/R
I=10/80
I=1/8 A

v) 0.02A? since current in the centre branch when V1 is short circuited is 0A.

for part V yes.

I also just re-read your origonal questions and realized I missed something for parts 2 and 4. You know the current through the source V1 due to V1, however the will be current flowing through the source V1 that is due to the source V2. the same is true for part 4.

 

[johndough999]

for part V yes.

I also just re-read your origonal questions and realized I missed something for parts 2 and 4. You know the current through the source V1 due to V1, however the will be current flowing through the source V1 that is due to the source V2. the same is true for part 4.

Does that mean the answers are incorrect?

 

[donpacino]

Does that mean the answers are incorrect?

Yes, you have half of the answer for parts two and four. You then need to use superposition to find the correct answer (hint, you already did most of the work)

 

[johndough999]

Yes, you have half of the answer for parts two and four. You then need to use superposition to find the correct answer (hint, you already did most of the work)

Why can i not just simply use ohms law to find the current for part 2 and 4?

 

[donpacino]

Why can i not just simply use ohms law to find the current for part 2 and 4?

You have to include the current from both sources

 

[johndough999]

You have to include the current from both sources

The same way as for the last part, where I find the current in the center branch?

 

[donpacino]

The same way as for the last part, where I find the current in the center branch?

Its the current due to V1 + the current due to V2. You had it correct

 

[johndough999]

Its the current due to V1 + the current due to V2. You had it correct

Thanks for the help!

 

[donpacino]

no problem